A mass attached to the free end of an ideal spring executes SHM according to the equation x = 0.5 sin(20 t) where x is in meters and t is in seconds. What is the magnitude of the maximum acceleration for this mass? A) 200 m/s² B) 10 m/s² C) 100 m/s² D) 20 m/s² E) (20/ $\pi$) m/s²

Question

Quizplus · Accepted Answer

The maximum acceleration in SHM is given by $a_{max} = \omega^2 \times A$, where $\omega$ is the angular frequency and $A$ is the amplitude. Here, $\omega = 20$ rad/s and $A = 0.5$ m. Thus, $a_{max} = 20^2 \times 0.5 = 400 \times 0.5 = 200$ m/s$^2$.

A Mass Attached to the Free End of an Ideal π\piπ

A Mass Attached to the Free End of an Ideal $\pi$